## Search found 48 matches

- Fri Dec 16, 2016 9:36 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2016 Secondary 3: Weird angle condition
- Replies:
**3** - Views:
**1922**

### Re: BdMO National 2016 Secondary 3: Weird angle condition

A synthetic solution : Let $ Q$ be a point such that $\triangle PQB $ is equilateral ,$C$ and $Q$ are on the same side of $PB$.Then $Q$ is the center of $\odot BPC$. So, $AQ$ is the perpendicular bisector of $BC$ . As,$\angle APB=150^{\circ}$, so $AP \perp QB \Rightarrow AP$ bisect $\angle QAB$ .So ...

- Sat Dec 10, 2016 11:49 am
- Forum: Number Theory
- Topic: USAJMO 2016
- Replies:
**1** - Views:
**1571**

- Tue Nov 29, 2016 11:38 pm
- Forum: Number Theory
- Topic: Infinitely many primes divide $1!+2!+\cdots +n!$
- Replies:
**1** - Views:
**1197**

### Re: Infinitely many primes divide $1!+2!+\cdots +n!$

Let's prove a more general result: For a prime $P$ ,define $C_p = \{x : P^x \mid S_n , $ for some $n \in \mathbb N \}$, then $C_P$ is a bounded. Proof : Assume , $C_P$ is not bounded for a prime $P$. Then - Claim 1 : $P^x \mid S_{px}$ ,for any $x \in \mathbb N $ Proof: As $C_P$ is unbounded ,there e...

- Tue Nov 22, 2016 4:10 pm
- Forum: Geometry
- Topic: APMO 2013 P5
- Replies:
**3** - Views:
**1813**

- Tue Nov 22, 2016 12:54 am
- Forum: Geometry
- Topic: A Beauty from Evan Chen
- Replies:
**1** - Views:
**1231**

### Re: A Beauty from Evan Chen

Let $AG$ and $MO$ meet $BC$ at $K,S$ respectively .$KM$ meets $PN$ at $R$. As $M$ is the center of $\gamma$,so $MO$ is the perpendicular bisector of $AG$ . $ \angle PAM =\angle PGM =90^o \Rightarrow AP\| BC \Rightarrow \angle APM =\angle OSN.$ $ON=\frac {1}{2} AH =AM $ and $\angle ONS =90^o =\angle ...

- Mon Nov 21, 2016 11:44 pm
- Forum: Geometry
- Topic: Two triangles and three collinear points
- Replies:
**1** - Views:
**1889**

### Re: Two triangles and three collinear points

$AD$,$BE$ and $CF$ concur on the orthocenter $H$ of $\triangle DEF$. Let $D_0,E_0,F_0 $ be the projection of $D,E,F$ on $EF,FD $ and $DE$ respectively .Let $\alpha$ be the circumcircle of $\triangle DEF$. Lemma : The perpendiculars to $BC, CA, AB$ through $ R,S,T $ respectively intersect at $H$ Proo...

- Sat Oct 15, 2016 7:06 pm
- Forum: Algebra
- Topic: Polynomials without real solutions
- Replies:
**1** - Views:
**2026**

### Re: Polynomials without real solutions

My solution: Let $P$ and $Q$ be the graph of $P(X),Q(X)$. $P,Q$ doesn't intersect (as $P(X)$ can't be equal to $Q(x)$).Assume $P(P(X))=Q(Q(X))$.Consider the points $A(P(X),P(P(X))),B(P(X),Q(P(X))),C(Q(X),P(Q(X))),D(Q(X),Q(Q(X)))$. $AD \parallel BC \parallel XX^{'}$ and $AB \parallel YY^{'} \paralle...

- Sun Sep 11, 2016 11:44 pm
- Forum: Geometry
- Topic: Cyclic intersections of $AH$ and $BO$ gives similar $\Delta$
- Replies:
**2** - Views:
**1359**

### Re: Cyclic intersections of $AH$ and $BO$ gives similar $\De

$\angle XAC =\angle OAB=\angle ABX$ .so $\odot AXB$ touches $AC$ at $A$ .Similarly $\odot BYC$, $\odot AZC$ touches $AB$ & $BC$ at $B$ & $C$ respectively .By Miquel's theorem $\odot AXB$ , $\odot BYC$ & $\odot AZC$ pass though a common point, say $P$ .(Using directed angles) $ \angle XPY = \angle XP...